A) \[(n-1){{(}^{2n}}{{C}_{n}})\]
B) \[n{{(}^{2n}}{{C}_{n}})\]
C) \[(n+1){{(}^{2n}}{{C}_{n}})\]
D) \[(n+1){{(}^{n}}{{C}_{n/2}})\]
Correct Answer: C
Solution :
[c] \[{{S}_{n}}={{a}_{0}}{{C}_{0}}^{2}+{{a}_{1}}{{C}_{1}}^{2}+{{a}_{2}}{{C}_{2}}^{2}+.......+{{a}_{n}}{{C}_{n}}^{2}\] \[\frac{{{S}_{n}}={{a}_{n}}{{C}_{n}}^{2}+{{a}_{n-1}}{{C}_{n-1}}^{2}+{{a}_{n-2}}^{2}+....+{{a}_{0}}{{C}_{0}}^{2}}{2{{S}_{n}}=({{a}_{0}}+{{a}_{n}}){{C}_{0}}^{2}+({{a}_{1}}+{{a}_{n-1}}){{C}_{1}}^{2}+...+({{a}_{n}}+{{a}_{0}}){{C}_{n}}^{1}}\]\[=(2n+2)({{C}_{0}}^{2}+{{C}_{1}}^{2}+{{C}_{2}}^{2}+.....+{{C}_{n}}^{2})\] \[\therefore \,\,\,\,{{S}_{n}}={{(n+1)}^{2n}}{{C}_{n}}\] \[[\because \,\,\,\,\,{{a}_{0}}+{{a}_{n}}={{a}_{1}}+{{a}_{n-1}}+.......=2n+2]\]You need to login to perform this action.
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